3rd Asian Pacific Mathematics Olympiad 1991 Problems

3rd Asian Pacific Mathematics Olympiad 1991 Problems

A1.  ABC is a triangle. G is the centroid. The line parallel to BC through G meets AB at B' and AC at C'. Let A'' be the midpoint of BC, C'' the intersection of B'C and BG, and B'' the intersection of C'B and CG. Prove that A''B''C'' is similar to ABC.

A2.  There are 997 points in the plane. Show that they have at least 1991 distinct midpoints. Is it possible to have exactly 1991 midpoints?

A3.  x_i and y_i are positive reals with ∑₁ⁿ x_i = ∑₁ⁿ y_i. Show that ∑₁ⁿ x_i²/(x_i + y_i) ≥ (∑₁ⁿ x_i)/2.

A4.  A sequence of values in the range 0, 1, 2, ... , k-1 is defined as follows: a₁ = 1, a_n = a_n-1 + n (mod k). For which k does the sequence assume all k possible values?

A5.  Circles C and C' both touch the line AB at B. Show how to construct all possible circles which touch C and C' and pass through A.

Solutions

ABC is a triangle. G is the centroid. The line parallel to BC through G meets AB at B' and AC at C'. Let A'' be the midpoint of BC, C'' the intersection of B'C and BG, and B'' the intersection of C'B and CG. Prove that A''B''C'' is similar to ABC.

Solution

Let M be the midpoint of AB and N the midpoint of AC. Let A''M meet BG at X. Then X must be the midpoint of A''M (an expansion by a factor 2 center B takes A''M to CA and X to N). Also BX/BN = 1/2 and BG/BN = 2/3, so XG = BX/3. Let the ray CX meet AB at Z. Then ZX = CX/3. (There must be a neat geometric argument for this, but if we take vectors origin B, then BX = BN/2 = BA/4 + BC/4, so BZ = BA/3 and so XZ = 1/3 (BA/4 - 3BC/4) = CX/3.) So now triangles BXC and ZXG are similar, so ZG is parallel to BC, so Z is B' and X is C''. But A''X is parallel to AC and 1/4 its length, so A''C'' is parallel to AC and 1/4 its length. Similarly A''B'' is parallel to AB and 1/4 its length. Hence A''B''C'' is similar to ABC. 

There are 997 points in the plane. Show that they have at least 1991 distinct midpoints. Is it possible to have exactly 1991 midpoints?

Solution

Answer: yes. Take the 997 points collinear at coordinates x = 1, 3, ... , 1993. The midpoints are 2, 3, 4, ... , 1992.

Take two points A and B which are the maximum distance apart. Now consider the following midpoints: M, the midpoint of AB, the midpoint of each AX for any other X in the set (not A or B), and the midpoint of each BX. We claim that all these are distinct. Suppose X and Y are two other points (apart from A and B). Clearly the midpoints of AX and AY must be distinct (otherwise X and Y would coincide). Similarly the midpoints of BX and BY must be distinct. Equally, the midpoint of AX cannot be M (or X would coincide with B), nor can the midpoint of BX be M. Suppose, finally, that N is the midpoint of AX and BY. Then AYXB is a parallelogram and either AX or BY must exceed AB, contradicting the maximality of AB. So we have found 1991 distinct midpoints. The example above shows that there can be exactly 1991 midpoints. 

x_i and y_i are positive reals with ∑₁ⁿ x_i = ∑₁ⁿ y_i. Show that ∑₁ⁿ x_i²/(x_i + y_i) ≥ (∑₁ⁿ x_i)/2.

Solution

We use Cauchy-Schwartz: ∑ (x/√(x+y) )² ∑ (√(x+y) )² ≥ (∑ x )². So ∑ x²/(x+y) >= (∑ x)²/(∑(x+y) = 1/2 ∑ x.

A sequence of values in the range 0, 1, 2, ... , k-1 is defined as follows: a₁ = 1, a_n = a_n-1 + n (mod k). For which k does the sequence assume all k possible values?

Solution

Let f(n) = n(n+1)/2, so a_n = f(n) mod k. If k is odd, then f(n+k) = f(n) mod k, so the sequence can only assume all possible values if a₁, ... , a_k are all distinct. But f(k-n) = f(n) mod k, so there are at most (k+1)/2 distinct values. Thus k odd does not work.

If k is even, then f(n+2k) = f(n) mod k, so we need only look at the first 2k values. But f((2k-1-n) = f(n) mod k and f(2k-1) = 0 mod k, so the sequence assumes all values iff a₁, a₂, ... , a_k-1 assume all the values 1, 2, ... , k-1.

Checking the first few, we find k = 2, 4, 8, 16 work and k = 6, 10, 12, 14 do not. So this suggests that k must be a power of 2. Suppose k is a power of 2. If f(r) = f(s) mod k for some 0 < r, s < k, then (r - s)(r + s + 1) = 0 mod k. But each factor is < k, so neither can be divisible by k. Hence both must be even. But that is impossible (because their sum is 2r+1 which is odd), so each of f(1), f(2), ... , f(k-1) must be distinct residues mod k. Obviously none can be 0 mod k (since 2k cannot divide r(r+1) for 0 < r < k and so k cannot divide f(r) ). Thus they must include all the residues 1, 2, ... k-1. So k a power of 2 does work.

Now suppose h divides k and k works. If f(n) = a mod k, then f(n) = a mod h, so h must also work. Since odd numbers do not work, that implies that k cannot have any odd factors. So if k works it must be a power of 2. 

Circles C and C' both touch the line AB at B. Show how to construct all possible circles which touch C and C' and pass through A.

Solution

Take a common tangent touching C' at Q' and C at Q. Let the line from Q to A meet C again at P. Let the line from Q' to A meet C' again at P'. Let the C have center O and C' have center O'. Let the lines OP and O'P' meet at X. Take X as the center of the required circle. There are two common tangents, so this gives two circles, one enclosing C and C' and one not.

To see that this construction works, invert wrt the circle on center A through B. C and C' go to themselves under the inversion. The common tangent goes to a circle through A touching C and C'. Hence the point at which it touches C must be P and the point at which it touches C' must be P'.